Question

plz solve this question
if you know
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Answer 1

Answer:

I don't know sorry

Answer 2

Answer:

Option a

Step-by-step explanation:

Given :-

In ∆ABC, DE||BC and AD = 4x-3 ,AE = 8x-7 ,

BD = 3x-1 and CE = 5x-3

To find :-

Find the value of x ?

Solution :-

Given that

In ∆ABC, D and E are the points on AB and AC respectively.

DE || BC

AD = 4x-3

AE = 8x-7

BD = 3x-1

CE = 5x-3

We know that

By Basic Proportionality Theorem,

AD/DB = AE/EC

On substituting these values in the above condition

=> (4x-3)/(3x-1) = (8x-7)/(5x-3)

On applying cross multiplication then

=> (4x-3)(5x-3) = (3x-1)(8x-7)

=> 4x(5x-3)-3(5x-3) = 3x(8x-7)-1(8x-7)

=> 20x²-12x-15x+9 = 24x²-21x-8x+7

=> 20x²-27x+9 = 24x²-29x+7

=> 24x²-29x+7-20x²+27x-9 = 0

=> 4x²-2x-2 = 0

=> 2(2x²-x-1) = 0

=> 2x²-x-1 = 0/2

=> 2x²-x-1 = 0

=> 2x²-2x+x-1 = 0

=> 2x(x-1)+1(x-1) = 0

=> (x-1)(2x+1) = 0

=> x-1 = 0 (or) 2x+1 = 0

=> x = 1 (or) 2x = -1

=> x = 1 (or) x = -1/2

But x can't be negative.

Therefore, x = 1

Answer:-

The value of x for the given problem is 1

Used Theorem:-

Basic Proportionality Theorem:-

(Thales Theorem)

" A line drawn parallel to the one side of a triangle intersecting other two sides at two different points then the line divides the other two sides in the sane ratio".